Answer
See the detailed answer below.
Work Step by Step
We can use the double-slit analyze here since the author told us to assume that all the measurements are made such farther than 9 meters from the anttenas (the anttenas here are representing the double-slit).
For maximum signals, which is the constructive interference, we know that
$$m\lambda=d\sin\theta$$
Thus,
$$\theta=\sin^{-1}\left[\dfrac{m\lambda}{d}\right]$$
where $m=0,1,2,3,...$
Now we need to find the wavelength since we know its frequency. The wavelength is given by
$$\overbrace{v}^{c}=\lambda f$$ so
$$\lambda=\dfrac{c}{f}$$
Plugging into the angle $\theta$ formula above
$$\theta=\sin^{-1}\left[\dfrac{mc}{fd}\right]$$
So that, the maximum angles are;
- When $m=0$;
$$\theta_1=\sin^{-1}\left[\dfrac{0\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{red}{\bf 0^\circ} $$
- When $m=1$;
$$\theta_2=\sin^{-1}\left[\dfrac{1\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{red}{\bf 21.66^\circ} $$
- When $m=2$;
$$\theta_3=\sin^{-1}\left[\dfrac{2\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{red}{\bf 47.6^\circ} $$
- When $m=3$;
$$\theta_4=\sin^{-1}\left[\dfrac{3\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{blue}{\bf undefined} $$
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For minimum signals, which is the destructive interference, we know that
$$\left(m+\frac{1}{2}\right)\lambda=d\sin\theta$$
Thus,
$$\theta=\sin^{-1}\left[\dfrac{\left(m+\frac{1}{2}\right)\lambda}{d}\right]$$
where $m=0,1,2,3,...$
Plugging $\lambda$ from above;
$$\theta=\sin^{-1}\left[\dfrac{\left(m+\frac{1}{2}\right)c}{fd}\right]$$
So that, the minimum angles are;
- When $m=0$;
$$\theta_1=\sin^{-1}\left[\dfrac{\left(0+\frac{1}{2}\right)\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{red}{\bf 10.64^\circ} $$
- When $m=1$;
$$\theta_2=\sin^{-1}\left[\dfrac{\left(1+\frac{1}{2}\right)\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{red}{\bf 33.62^\circ} $$
- When $m=2$;
$$\theta_3=\sin^{-1}\left[\dfrac{\left(2+\frac{1}{2}\right)\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{red}{\bf 67.35^\circ} $$
- When $m=3$;
$$\theta_4=\sin^{-1}\left[\dfrac{\left(3+\frac{1}{2}\right)\times 3\times10^8 }{90.3\times10^6\times 9}\right]=\color{blue}{\bf undefined} $$
